Question

A sample of 20 different iPads is randomly selected from a production batch containing 3% defectives.

What is the probability that exactly 4 are working properly?

What is the probability that at most one from the 20 randomly selected is defective?

Answer 1

Solution

Given that ,

a)

p = 0.97

1 - p = 0.03

n = 20

x = 4

Using binomial probability formula ,

P(X = x) = ((n! / x! (n - x)!) * px * (1 - p)n - x

P(X = 4) = ((20! / 4! (20 - 4)!) * 0.974 * (0.03)20 - 4

= 0

Probability = 0.000

b)

Given that ,

p = 0.03

1 - p = 0.97

n = 20

Using binomial probability formula ,

P(X = x) = ((n! / x! (n - x)!) * px * (1 - p)n - x

P(X 1) = P(X = 0) + P(X = 1)

= ((20! / 0! (20 - 0)!) * 0.030 * (0.97)20 - 0 + ((20! / 1! (20 - 1)!) * 0.031 * (0.97)20 - 1

= 0 + 0

Probability = 0.000